Question: Solve for $y$, $ -\dfrac{3y - 3}{y - 4} = -\dfrac{5}{3y - 12} + \dfrac{1}{5y - 20} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $y - 4$ $3y - 12$ and $5y - 20$ The common denominator is $15y - 60$ To get $15y - 60$ in the denominator of the first term, multiply it by $\frac{15}{15}$ $ -\dfrac{3y - 3}{y - 4} \times \dfrac{15}{15} = -\dfrac{45y - 45}{15y - 60} $ To get $15y - 60$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{5}{3y - 12} \times \dfrac{5}{5} = -\dfrac{25}{15y - 60} $ To get $15y - 60$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{1}{5y - 20} \times \dfrac{3}{3} = \dfrac{3}{15y - 60} $ This give us: $ -\dfrac{45y - 45}{15y - 60} = -\dfrac{25}{15y - 60} + \dfrac{3}{15y - 60} $ If we multiply both sides of the equation by $15y - 60$ , we get: $ -45y + 45 = -25 + 3$ $ -45y + 45 = -22$ $ -45y = -67 $ $ y = \dfrac{67}{45}$